Assign oxidation number to the underlined elements in each of the following species:
H4P2O7
Let the oxidation number(O.N.) of P be ‘x’
We know that,
O.N. of H = + 1
O.N. of O = -2
Therefore, we have,
4(+ 1) + 2(x) + 7(-2) = 0
4 + 2x-14 = 0
X = + 5
The O.N. of P is + 5.